Question

find the standard form for the equation of a circle $(x - h)^2+(y - k)^2=r^2$ with a diameter that has endpoints $(-7,10)$ and $(9,-5)$. $h =$ $k =$ $r =$ question help: video 1 video 2 check answer

Explanation:

Step1: Find the center \((h,k)\)

The center of the circle is the mid - point of the diameter. The mid - point formula is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).
For the points \((x_1=-7,y_1 = 10)\) and \((x_2=9,y_2=-5)\), we have:
\(h=\frac{-7 + 9}{2}=\frac{2}{2}=1\)
\(k=\frac{10+( - 5)}{2}=\frac{10 - 5}{2}=\frac{5}{2}=2.5\)

Step2: Find the radius \(r\)

The radius is half the length of the diameter. The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
First, find the length of the diameter:
\(d=\sqrt{(9-( - 7))^2+(-5 - 10)^2}=\sqrt{(9 + 7)^2+(-15)^2}=\sqrt{16^2+(-15)^2}=\sqrt{256 + 225}=\sqrt{481}\)
Then \(r=\frac{\sqrt{481}}{2}\)

Answer:

\(h = 1\)
\(k=2.5\)
\(r=\frac{\sqrt{481}}{2}\)